How do you solve this?

[Lovely918]

x dy/dx +y-e^x= 0 What method would you go about by solving this? WOuld you use dy/dx + Py = Q?

 

[Deleted member 4993]

x dy/dx +y-e^x= 0 What method would you go about by solving this? WOuld you use dy/dx + Py = Q?

Yes - and the integrating factor would be \(\displaystyle e^{\int P(x)dx}\) where P(x) =\(\displaystyle \dfrac{1}{x}\) and Q(x) = \(\displaystyle \dfrac{e^x}{x}\)

 

[Lovely918]

Thanks!

Thank you !!!!

Yes - and the integrating factor would be \(\displaystyle e^{\int P(x)dx}\) where P(x) =\(\displaystyle \dfrac{1}{x}\) and Q(x) = \(\displaystyle \dfrac{e^x}{x}\)

 

[Lovely918]

Hmmm.

If the question is the integral of px then what part does qx play?

 

[Deleted member 4993]

If the question is the integral of px then what part does qx play?

\(\displaystyle y' + P(x)*y = Q(x)\)


\(\displaystyle y' * e^{\int P(x) dx} \ + \ P(x) \ * \ y \ * \ e^{\int P(x) dx} \ = \ Q(x) \ * \ \ e^{\int P(x) dx} \ \)


\(\displaystyle \dfrac {d}{dx}\left [ y \ * \ e^{\int P(x) dx} \right ] \ = \ Q(x) \ * \ \ e^{\int P(x) dx} \ \)

Now integrate both sides - and solve for 'y'

 

[Lovely918]

So is it like this?

so do you solve it like this:

d/dx (y*e^ integral 1/x dx)=((e^x)/x)* e^(integral 1/x dx)


Is that right?

 

[mmm4444bot]

If the question is the integral of px then what part does qx play?

There is no px or qx in this exercise, and finding the integral of function P is not the question.

This exercise wants you to find some function y = f(x) = ???? such that function f satisfies the given differential equation.

Cheers :cool:

 

[Deleted member 4993]

so do you solve it like this:

d/dx (y*e^ integral 1/x dx)=((e^x)/x)* e^(integral 1/x dx)


Is that right?

Right - but final answer has not been reached.

 

[mmm4444bot]

so do you solve it like this:

d/dx (y*e^ integral 1/x dx)=((e^x)/x)* e^(integral 1/x dx)


Is that right?

I may be misinterpreting your questions above.

I can tell you that you have correctly substituted the algebraic definitions that Subhotosh supplied for functions P and Q into the equation provided to you in Subhotosh's last post.

 

[HallsofIvy]

so do you solve it like this:

d/dx (y*e^ integral 1/x dx)=((e^x)/x)* e^(integral 1/x dx)


Is that right?

Do you not know what \(\displaystyle \int 1/x dx\) is? If you do, why have you not put it in? This is a very very simple integration problem.

Also, I would like to point out that, by the product rule, \(\displaystyle \frac{d(xy)}{dx}= \frac{dx}{dx}y+ x\frac{dy}{dx}= x\frac{dy}{dx}+ y\)

So, your differential equation is \(\displaystyle x\frac{dy}{dx}+ y= \frac{d(xy)}{dx}= e^x\). That is what you would get using the more "formal" integrating factor method.